Tag: quantum physics

Questions Related to quantum physics

The photoelectric cut off voltage in a certain experiment is 1.5 V. The maximum kinetic energy of photoelectrons emitted is then

  1. 2.4 eV

  2. 1.5 eV

  3. 3.1 eV

  4. 4.5 eV


Correct Option: B
Explanation:

The minimum negative potential applied to the plate or anode  for which the photoelectric current just becomes zero, So, in this case, maximum K.E of an electron will be equal to stopping potential.
Here, $V _0 \, = \, 1.5 \, V,$
Maximum Kinetic energy = $eV _{0} \, = \, 1.5 \,eV$

Light of wavelength 0.6 mm from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5 V. With light of wavelength 0.4 mm from a sodium lamp, the stopping potential is 1.5 V. With this data, the value of h/e is:

  1. $6\times {{10}^{-5}}\,V{{s}^{-1}} $

  2. $2\times {{10}^{-15}}\,V{{s}^{-1}} $

  3. $4\times {{10}^{-55}}\,V{{s}^{-1}} $

  4. $4\times {{10}^{-15}}\,V{{s}^{-1}} $


Correct Option: D
Explanation:

In emission of electron, Potential Energy

$eV = \dfrac{hc}{\lambda} - W _0$

When light of wavelength $\lambda =0.6\,mm$and stopping potential $0.5\,V$
$0.5e=\dfrac{hc}{6\times {{10}^{-7}}}-{{W} _{0}}\ ......\ (1)$

When light of wavelength $\lambda =0.4\,mm$and stopping potential $1.5\,V$
$1.5e=\dfrac{hc}{4\times {{10}^{-7}}}-{{W} _{0}}\ ......\ (2)$
subtract equation (1) from (2) $ e=\dfrac{hc}{{{10}^{-7}}}\left[ \dfrac{1}{4}-\dfrac{1}{6} \right] $

$ \Rightarrow \dfrac{h}{e}=\dfrac{12\times {{10}^{-7}}}{3\times {{10}^{8}}}=4\times {{10}^{-15}}\,V{{s}^{-1}} $ 

Photons absorbed in meter are converted to heat. A source emitting $n$ photons/s of frequency $v$ is used to convert $1\ kg$ of ice of ${0}^{o}c$ to water at ${0}^{o}C$. Then, the time taken for the conversion:

  1. decreases with increasing $n$, with $v$ fixed

  2. decreases with $n$ fixed, $v$ increasing

  3. remains constant with $n$ and $v$ changing such that $nv=$constant

  4. increases when the product $nv$ increases


Correct Option: A

The energy od proton in a light of wavelength 6000 $\mathring { A } $ is :
$(h = 6.63 \times 10^{-34}$ $Js,$ $c = 3 \times 10^8 m/s)$

  1. $6.81 \times 10^{-19} J$

  2. $5.61 \times 10^{-19} J$

  3. $3.31 \times 10^{-19} J$

  4. $2.31 \times 10^{-19} J$


Correct Option: C
Explanation:

$E = {{hc} \over \alpha }$

$ = {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {6000 \times {{10}^{ - 10}}}}$

$ = 3.31 \times {10^{ - 19}}J$

A photo-electric threshold wavelength of tungsten is $2300 \mathring{A}$ . The energy of electrons ejected from the surface, if ultra -violet light of wavelength $1800 \mathring{A}$ is incident on it, is 

  1. 1.5 eV

  2. 2 eV

  3. 3.2 eV

  4. 6 eV


Correct Option: C

Calculate the kinetic energy of the electron having wavelength 1 nm

  1. 2.1 eV

  2. 3.1 eV

  3. 1.5 eV

  4. 4.2 eV


Correct Option: C

A laser beam ($\iota = 633\ nm$)has an power of $3mW$. What will be the pressure exerted on a surface by this beam if the cross sectional area is $3\ mm^{2}$ (Assume perfect reflection and normal incidence)

  1. $6.6\times 10^{-3}N/m^{2}$

  2. $6.6\times 10^{-6}N/m^{2}$

  3. $6.6\times 10^{-9}N/m^{2}$

  4. $6.6\ N/m^{2}$


Correct Option: B

A metalic surface is irradiated with monochr matic light of variable wavelength. Above wavelength of $ 5000 \dot { A }  $ , no photoelectrons a emitted from the surface. With an unknown wavelength, a stopping potential of 3V is nessary to eliminate the photo current . The Known wavelength is:

  1. 2258 $ \dot { A } $

  2. $ 4133 \dot { A } $

  3. $ 3126 \dot { A } $

  4. $ 2679 \dot { A } $


Correct Option: A

Two free protons are separated by a distance of $1\mathring A $. If they are released the kinetic energy of proton when it infinite separation is

  1. $23 \times 10 ^ { 19 } j$

  2. $11.5 \times 10 ^ { - 19 } \mathrm { J }$

  3. $46 \times 10 ^ { - 19 } \mathrm { J }$

  4. $5.6 \times 10$


Correct Option: B

The maximum energy of emitted photo electrons is measured by

  1. The current they produce

  2. The potential difference they produce

  3. The largest potential difference they can traverse

  4. The speed with which they emerge


Correct Option: C
Explanation:

Maximum energy of emitted photo electrons is equal to the stopping potential. Stopping potential is the largest potential difference an emitted photo electron can traverse.